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2 Answers

Shikhar Vishnoi

11 points Registered user

1

I think it would be : (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))9/10 takes care of leading 0 ( since all the digits have equal chance of being leading digit10C5 selects first 5 different digits5C2 selects 2 different digits from already used 5 digits to make the number look like (aabbcde) 7!/2!2! is the number of permutations for such numbers (aabbcde)5C1 selects 1 digit from already used 5 digits to make the number look like (aaabcde)7!/3! is the number of permutations for such numbers(aaabcde)hence the final answer is (9/10)*10C5*(5C2*(7!/(2!*2!))+5C1*(7!/3!))